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Is an an easy way to see that if $f: U \subset \mathbb R^n \rightarrow V \subset \mathbb R^m$ and $g: V \subset \mathbb R^m \rightarrow \mathbb R^p$ are smooth functions then their composite $g \circ f: U \rightarrow \mathbb R^p$ is smooth ?

I've been looking at the definiton of the derivative (multivariable) and the chain rule. However, help is needed in order to fully see that $g \circ f$ is smooth.

Looking at the one-dimensional case, I see easily that $g \circ f$ is smooth by elementary rules for differentiation.

Shuzheng
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1 Answers1

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If you want hard proof, you're basically ask for a proof of the chain rule. There are many proofs of this and it's not necessary to post one on MSE. Though, if you want help with a particular step in a proof you have you could post that as your question. If intuition is what you ask for, then think of smooth functions as just that, a transformation that neither tears nor creases. If you do one of these, and then another, you still won't be tearing or creasing.

Edit: (In light of the comment)

You need to specify more than "smooth", if you really are asking about higher order derivatives, as it has more than one usage. I'll describe and prove a statement similar to what I think you want, in the case that $f,g$ are both $\mathcal C^{\infty}$. You are right in that the bare chain rule does not state this.

Proposition: If $f, g$ (as you described) are $\mathcal C^{\infty}$ functions, then so is $g\circ f$.

Proof: We need to prove that each partial derivative of $h =g\circ f$ is $\mathcal C^{\infty }$. With respect to standard cartesian coordinates, the chain rule says that $\frac{\partial h_i}{\partial x_j}$ exists for each $1\leq i \leq p$, $1\leq j \leq n$, and is equal to $$\frac{\partial h_i}{\partial x_j} = \sum_{k=1}^m\left(\frac{\partial g_i}{\partial x_k}\circ f\right)\cdot\frac{\partial f_k}{\partial x_j}$$ But all of the partial derivatives of $f$ and of $g$ are $\mathcal C^{\infty}$ by assumption, and we have $\frac{\partial h_i}{\partial x_j}$ expressed as products, sums, and composition with $f$ of these, which we know maintain differentiability. Therefore $h$ is $\mathcal C^{\infty}$.

In the case you only assume $f$ or $g$ are only $\mathcal C^r$, $\mathcal C^s$, respectively, then $h$ will have the least amount of differentiability of the two. For example, suppose $1\leq r\leq s$. Then $h$ will be $\mathcal C^r$, and you can come up with an example where an $(r+1)^{\text{}th}$ partial derivative of $h$ is not continuous (in one dimension this is easy).

GPerez
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  • But I think I understand the proof of the Chain Rule in multivariable calculus. But then I must prove that the second order, third order, ... derivatives exist. As I see it, this doesn't follow from the Chain Rule or does it ? I know $D(g \circ f) = D(g) \circ D(f)$. – Shuzheng Apr 13 '15 at 14:53
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    @user111854: If $f$ and $g$ are smooth functions, their partial derivatives will be smooth functions, so you can apply the chain rule again to what you got when you computed the first derivative, and this proves that the second derivative exists. And so on... – Hans Lundmark Apr 13 '15 at 15:12
  • @HansLundmark Your comment crystallized this for me and hopefully is what the OP needs as well. Thanks. – Todd Wilcox Apr 13 '15 at 17:08
  • Why does $f$ smooth imply partial derivatives of all orders smooth ? I've never seen such a theorem. – Shuzheng Apr 13 '15 at 17:15
  • And why does the chain rule imply that partial derivatives exist ? – Shuzheng Apr 13 '15 at 17:26
  • Smooth means that $f$ is differentiable of all orders $k \ge 1$ ? – Shuzheng Apr 13 '15 at 17:27
  • @user111854 There isn't a universal definition of "smooth". Depending on context it can be differentiable, differentiable with continuity a certain number of times, of infinitely differentiable. I took it to be $\mathcal C^{\infty}$ to simplify the statements, but I also outlined how to treat other cases. And the chain rule implies that the differential $D h$ exists, so in particular every directional derivative exists. – GPerez Apr 13 '15 at 17:33